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Article

Solving an Integral Equation via Tricomplex-Valued Controlled Metric Spaces

by
Rajagopalan Ramaswamy
1,*,
Gunaseelan Mani
2,
Arul Joseph Gnanaprakasam
3,
Ola A. Ashour Abdelnaby
1,4 and
Stojan Radenović
5
1
Department of Mathematics, College of Science and Humanities in Alkharj, Prince Sattam Bin Abdulaziz University, Al-Kharj 11942, Saudi Arabia
2
Department of Mathematics, Saveetha School of Engineering, Saveetha Institute of Medical and Technical Sciences, Chennai 602105, India
3
Department of Mathematics, College of Engineering and Technology, SRM Institute of Science and Technology, Kattankulathur, Chennai 603203, India
4
Department of Mathematics, Cairo University, Cairo 12613, Egypt
5
Faculty of Mechanical Engineering, University of Belgrade, Kraljice Marije 16, 11120 Belgrad, Serbia
*
Author to whom correspondence should be addressed.
Axioms 2023, 12(1), 56; https://doi.org/10.3390/axioms12010056
Submission received: 30 November 2022 / Revised: 25 December 2022 / Accepted: 28 December 2022 / Published: 3 January 2023
(This article belongs to the Special Issue Fixed Point Theory and Its Related Topics III)

Abstract

:
In this present study, we propose the concept of tricomplex-controlled metric spaces as a generalization of both controlled metric-type spaces and tricomplex metric-type spaces. In this work, we establish fixed point results using Banach, Kannan and Fisher-type contractions supported with nontrivial examples in the setting of the proposed space. We apply the derived result to find the analytical solution of an integral equation using the fixed point technique under the same metric.

1. Introduction

In 1892, Segre published the work Real Representations of Complex Forms and Hyper Algebraic Bodies [1], in which he inserted the geometrical interpretation of the algebra of bicomplex numbers, returning after forty years to the interrupted thread of Hamilton’s thought. Segre introduced bicomplex points as a natural completion of the complex projective straight line, as well as bicomplex numbers with hyperalgebraic entities (complex entities that are algebraic in a real representation). The commutative generalization of complex numbers is conceptualised as bcn (briefly bicomplex numbers), tcn (briefly tricomplex numbers), etc., as elements of an infinite set of algebra. Subsequently, between 1930–1940, other researchers also contributed in this area. Readers are invited to refer to the works [2,3,4]. However, unfortunately, the next fifty years failed to witness any advancement in this field.
Later, Price [5] introduced the singularities of holomorphic functions of bicomplex variables (An introduction to Multicomplex Spaces and Functions, Dekker, New York, 1991). Recently, renewed interest in this subject has found some significant applications in different fields of mathematical sciences, as well as other branches of science and technology. Fabrizio Colombo, Irene Sabadini, Daniele Struppa, Adrian Vajiac and Mihaela Vajiac [6] established that even in the case of one bicomplex variable there cannot be compact singularities, with the help of computational algebraic techniques. These techniques allow us to prove the duality theorem for such functions. Many researchers have reported their findings in this arena.
Among them, in 2012, Luna-Elizaarrarás et al. [7] introduced the algebra of bicomplex numbers as a generalization of the field of complex numbers and described how to define elementary functions such as the polynomial, exponential and trigonometric functions in such an algebra, as well as their inverse functions (roots, logarithms and inverse trigonometric functions). Later. in 2017, Choi et al. [8] established fixed point results with two weakly compatible mappings in the setting of bicomplex-valued metric spaces using the E.A. Property. In the same year, Dhivya et al. [9] reported fixed point results using rational contractions in ordered complex partial metric spaces. In 2019, Jebril [10] proved some common fixed point theorems using rational contractions for a pair of mappings in bicomplex-valued metric spaces.
In 2018, Mlaiki et al. [11] presented the concept of controlled metric and established fixed point results in the setting of these spaces. For some more results on these spaces, readers are requested to refer to [12,13,14].
In 2021, Beg, Kumar Datta and Pal [15] proved FPT on bicomplex-valued metric spaces. In the sequel, Gunaseelan et al. [16], Aslam et al. [17] and Zhaohui et al. [18] studied the existence of unique solutions of nonlinear integral equations using fixed point results in the setting of complex-valued metric spaces and their extensions.
Recently, in 2022, Rajagopalan R. et al. [19] proved fixed point theorems on tricomplex metric spaces using control functions. They introduced a tricomplex-controlled metric space with mapping F : × × [ 0 , 1 ) , established an analogue of Banach-type contraction and applied the fixed point result to find a solution to an integral equation.
Inspired by this, in this paper, we introduce the concept of a tricomplex-controlled metric space with mapping F : × [ 1 , ) (in short, Tcvcms) and prove fixed point theorems using Banach-type, Kannan-type and Fisher-type contractions in the settings of these spaces.
The rest of the paper is organized as follows: In Section 2, we review some basic definitions from the literature and monograph, which are required in the sequel. In Section 3, we establish fixed point results in the setting of Tcvcms, using Banach- and Kannan-type contractions, supported with suitable examples for the derived results. In Section 4, we apply the derived result to find an analytical solution to integral equations. Finally, we conclude the article with some open problems to examine the possibility of extending/generalizing our results.

2. Preliminaries

Throughout this paper, we denote the set of real, complex, bicomplex and tricomplex numbers, respectively, as C 0 , C 1 , C 2 and C 3 . Price [5] defined the bcn as:
ζ = 1 + 2 i 1 + 3 i 2 + 4 i 1 i 2 ,
where 1 , 2 , 3 , 4 C 0 , and independent units i 1 , i 2 are such that i 1 2 = i 2 2 = 1 and i 1 i 2 = i 2 i 1 ; we denote the set of bcn C 2 as:
C 2 = { ζ : ζ = 1 + 2 i 1 + 3 i 2 + 4 i 1 i 2 , 1 , 2 , 3 , 4 C 0 } ,
i.e.,
C 2 = { ζ : ζ = κ 1 + i 2 κ 2 , κ 1 , κ 2 C 1 } ,
where κ 1 = 1 + 2 i 1 C 1 and κ 2 = 3 + 4 i 1 C 1 . Price [5] defined the tricomplex number as:
ϖ = 1 + 2 i 1 + 3 i 2 + 4 j 1 + 5 i 3 + 6 j 2 + 7 j 3 + 8 i 4 ,
where 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 C 0 , and independent units i 1 , i 2 , i 3 , i 4 , j 1 , j 2 , j 3 are such that i 1 2 = i 2 2 = i 3 2 = i 4 2 = 1 , i 4 = i 1 j 3 = i 1 i 2 i 3 , j 2 = i 1 i 3 = i 3 i 1 , j 2 2 = 1 , j 1 = i 1 i 2 = i 2 i 1 and j 1 2 = 1 .
We denote the set of tcn C 3 as:
C 3 = { ϖ : ϖ = 1 + 2 i 1 + 3 i 2 + 4 j 1 + 5 i 3 + 6 j 2 + 7 j 3 + 8 i 4 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 C 0 } ,
i.e.,
C 3 = { ϖ : ϖ = ζ 1 + i 3 ζ 2 , ζ 1 , ζ 2 C 2 } ,
where ζ 1 = κ 1 + κ 2 i 2 C 2 and ζ 2 = κ 3 + κ 4 i 2 C 2 , such that κ 3 = 5 + 6 i 1 and κ 4 = 7 + 8 i 1 . If ϖ = ζ 1 + i 3 ζ 2 and μ = 1 + i 3 2 are any two tcn, then the sum is ϖ ± μ = ( ζ 1 + i 3 ζ 2 ) ± ( 1 + i 3 2 ) = ζ 1 ± 1 + i 3 ( ζ 2 ± 2 ) , and the product is ϖ · μ = ( ζ 1 + i 3 ζ 2 ) ( 1 + i 3 2 ) = ( ζ 1 1 ζ 2 2 ) + i 3 ( ζ 1 2 + ζ 2 1 ) .
There are four idempotent elements in C 3 ; they are 0 , 1 , g 1 = 1 + j 3 2 , g 2 = 1 j 3 2 , out of which g 1 and g 2 are nontrivial, such that g 1 + g 2 = 1 and g 1 g 2 = 0 . Every tricomplex number ζ 1 + i 3 ζ 2 can be uniquely expressed as the combination of g 1 and g 2 , namely
ϖ = ζ 1 + i 3 ζ 2 = ( ζ 1 i 2 ζ 2 ) g 1 + ( ζ 1 + i 2 ζ 2 ) g 2 .
This representation of ϖ is known as the idempotent representation of tricomplex number, and the complex coefficients ϖ 1 = ( ζ 1 i 2 ζ 2 ) and ϖ 2 = ( ζ 1 + i 2 ζ 2 ) are known as idempotent components of the bcn ϖ .
An element ϖ = ζ 1 + i 3 ζ 2 C 3 is said to be invertible if there exists another element μ in C 3 such that ϖ μ = 1 and μ is said to be multiplicative inverse of ϖ . Consequently, ϖ is said to be the multiplicative inverse of μ . An element which has an inverse in C 3 is said to be the nonsingular element of C 3 , and an element which does not have an inverse in C 3 is said to be the singular element of C 3 .
An element ϖ = ζ 1 + i 3 ζ 2 C 3 is nonsingular if | ζ 1 2 + ζ 2 2 | 0 and singular if | ζ 1 2 + ζ 2 2 | = 0 .
The inverse of ϖ is defined as
ϖ 1 = μ = ζ 1 i 3 ζ 2 ζ 1 2 + ζ 2 2 .
The norm | | . | | of C 3 is a positive real-valued function and | | . | | : C 3 C 0 + is defined by
| | ϖ | | = | | ζ 1 + i 3 ζ 2 | | = { | ζ 1 | 2 + | ζ 2 | 2 } 1 2 = | ( ζ 1 i 2 ζ 2 ) | 2 + | ( ζ 1 + i 2 ζ 2 ) | 2 2 1 2 = ( 1 2 + 2 2 + 1 2 + 3 2 + 4 2 + 5 2 + 6 2 + 7 2 + 8 2 ) 1 2 ,
where ϖ = 1 + 2 i 1 + 3 i 2 + 4 j 1 + 5 i 3 + 6 j 2 + 7 j 3 + 8 i 4 = ζ 1 + i 3 ζ 2 C 3 . The normed linear space C 3 with respect to norm | | . | | is a Banach space, as it is complete. If ϖ , μ C 3 , then | | ϖ μ | | 2 | | ϖ | | | | μ | | holds instead of | | ϖ μ | | | | ϖ | | | | μ | | ; therefore, C 3 is not Banach algebra.
The partial order relation i 3 on C 3 is defined as follows:
Let C 3 be the set of tcn and ϖ = ζ 1 + i 3 ζ 2 , μ = 1 + i 3 2 C 3 , then ϖ i 3 μ if and only if ζ 1 i 2 1 and ζ 2 i 2 2 , i.e., ϖ i 3 μ , if one of the following conditions is fulfilled:
(a)
ζ 1 = 1 , ζ 2 = 2 ;
(b)
ζ 1 i 2 1 , ζ 2 = 2 ,;
(c)
ζ 1 = 1 , ζ 2 i 2 2 , and;
(d)
ζ 1 i 2 1 , ζ 2 i 2 . 2 .
In particular, we can write ϖ i 3 μ if ϖ i 3 μ and ϖ μ , i.e., one of (b), (c) and (d) holds, and we write ϖ i 3 μ if (d) holds.
For any two tcn ϖ , μ C 3 , we can verify the following:
(1)
ϖ i 3 μ iff | | ϖ | | | | μ | | ;
(2)
| | ϖ + μ | | | | ϖ | | + | | μ | | ;
(3)
| | ϖ | | = | | | | ϖ | | , where ♭ is a positive real number;
(4)
| | ϖ μ | | 2 | | ϖ | | | | μ | | and the equality holds only when at least one of ϖ and μ is nonsingular;
(5)
| | ϖ 1 | | = | | ϖ | | 1 if ϖ is a nonsingular;
(6)
| | ϖ μ | | = | | ϖ | | | | μ | | , if μ is a nonsingular.
Now, let us recall some basic concepts and notations which are used in the sequel.
Definition 1
([11]). Let and φ : × [ 1 , ) . The functional £ c m : × [ 0 , ) is said to be a controlled metric type if
(CTM1)
£ c m ( α , g ) = 0 α = g ;
(CTM2)
£ c m ( α , g ) = £ c m ( g , α ) ;
(CTM3)
£ c m ( α , ξ ) φ ( α , g ) £ c m ( α , g ) + φ ( g , ξ ) £ c m ( g , ξ ) ;
for all α , g , ξ . Then, the pair ( , £ c m ) is known as a controlled metric type space.
Definition 2.
Let and consider φ : × [ 1 , ) . The functional £ c v c m s : × C 3 is said to be Tcvcms if
(TCCMS1)
0 i 3 £ c v c m s ( α , g ) also £ c v c m s ( α , g ) = 0 α = g ;
(TCCMS2)
£ c v c m s ( α , g ) = £ c v c m s ( g , α ) ;
(TCCMS3)
£ c v c m s ( α , ξ ) i 3 φ ( α , g ) £ c v c m s ( α , g ) + φ ( g , ξ ) £ c v c m s ( g , ξ ) ;
for all α , g , ξ . Then, the pair ( , £ c v c m s ) is known as a Tcvcms.
Definition 3.
Consider ( , £ c v c m s ) is a Tcvcms with a sequence { α ρ } in and α . Then:
(i)
A sequence { α ρ } in is convergent and converges to α if, for every 0 i 3 C 3 , there exists a natural number N so that £ κ ( α ρ , α ) i 3 for every ρ N . Then, we say lim ρ α ρ = α or α ρ α as ρ .
(ii)
If, for every 0 i 3 where C 3 , there exists a natural number N so that £ κ ( α ρ , α ρ + θ ) i 3 for every θ N and ρ > N . Then, { α ρ } is known as a Cauchy sequence in ( , £ κ ) .
(iii)
Tcvcms ( , £ c v c m s ) is complete if every Cauchy sequence in ⋓ is convergent in .
Definition 4.
Let ( , £ c v c m s ) be a Tcvcms and ( k = 1 ρ α k g k ) 2 i 3 ( k = 1 ρ α k 2 ) ( k = 1 ρ g k 2 ) . Then, this inequality is called tricomplex Lagrange’s inequality.
Lemma 1.
Let ( , £ c v c m s ) be a Tcvcms. Then, a sequence { α ρ } in is Cauchy sequence, such that α θ α ρ , wherever θ ρ . Then, { α ρ } converges to at most one point.
Proof. 
Consider the sequence { α ρ } with two limit points α * and g * and lim ρ £ c v c m s ( α ρ , α * ) = 0 = £ c v c m s ( α ρ , g * ) . Since { α ρ } is Cauchy, from TCCMS 3 , for α θ α ρ , whenever θ ρ , we can write
| | £ c v c m s ( α * , g * ) | | ı ˙ 3 [ φ ( α * , α ρ ) | | £ c v c m s ( α * , α ρ ) | | + φ ( α ρ , g * ) | | £ c v c m s ( α ρ , g * ) | | ] 0 as ρ .
We obtain | | £ c v c m s ( α * , g * ) | | = 0 , i.e., α * = g * . Thus, { α ρ } converges to at most one point. □
Lemma 2.
For a given Tcvcms ( , £ c v c m s ) , the tricomplex-valued controlled metric function £ c v c m s : × C 3 is continuous with respect to the partial order “ i 3 ”.
Proof. 
Let s , r C 3 , such that s r , then we prove that the set
£ c v c m s 1 ( r , s ) : = { ( α , g ) × | r i 3 £ c v c m s ( α , g ) i 3 s } ,
is open in the product topology on × . A basis for the product topology is the collection of all Cartesian products of open balls in ( , £ c v c m s ) .
Let ( α , g ) £ c v c m s 1 ( r , s ) . We choose ϵ = 1 200 min ( £ c v c m s ( α , g ) r , s £ c v c m s ( α , g ) ) . Then, for any point ( ϱ , ζ ) β ( α , ϵ ) × β ( g , ϵ ) , we have
£ c v c m s ( ϱ , ζ ) i 3 £ c v c m s ( ζ , α ) + £ c v c m s ( α , g ) + £ c v c m s ( g , ζ ) i 3 2 ϵ + £ c v c m s ( α , g ) i 3 s
and
r i 3 £ c v c m s ( α , g ) 2 ϵ i 3 £ c v c m s ( ϱ , ζ ) + £ c v c m s ( α , ζ ) ϵ + £ c v c m s ( g , ζ ) ϵ i 3 £ c v c m s ( ϱ , ζ ) .
Then, ( α , g ) β ( x , ϵ ) × β ( g , ϵ ) £ c v c m s 1 ( r , s ) . □
Define Fix Γ : = { α * | α * = Γ ( α * ) } as the set of fixed points.
In the following section, we present our main results by proving fixed point results supported with suitable examples and applications to an integral equation.

3. Main Results

Now, we prove the Banach-type contraction principle in the setting of Tcvcms.
Theorem 1.
Let ( , £ c v c m s ) be a complete Tcvcms and Γ : be a continuous mapping such that
£ c v c m s ( Γ α , Γ g ) i 3 £ c v c m s ( α , g ) ,
for all α , g , where 0 < < 1 . For α 0 , we denote α ρ = Γ ρ α 0 . Suppose that
max θ 1 lim r φ ( α r + 1 , α r + 2 ) φ ( α r , α r + 1 ) φ ( α r + 1 , α θ ) < 1 .
In addition, for every α , if the limits
lim ρ φ ( α ρ , α ) and lim ρ φ ( α , α ρ ) exist and are finite ,
then, Γ has a UFP (briefly unique fixed point).
Proof. 
Consider the sequence { α ρ = Γ ρ α 0 } . From (1), we obtain
£ c v c m s ( α ρ , α ρ + 1 ) i 3 £ c v c m s ( α ρ 1 , α ρ ) i 3 . . . . i 3 ρ £ c v c m s ( α 0 , α 1 ) , ρ 0 .
For all ρ < θ , where ρ , θ N , we have
£ c v c m s ( α ρ , α θ ) i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + φ ( α ρ + 1 , α θ ) £ c v c m s ( α ρ + 1 , α θ ) i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 1 , α ρ + 2 ) £ c v c m s ( α ρ + 1 , α ρ + 2 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 2 , α θ ) £ c v c m s ( α ρ + 2 , α θ ) i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 1 , α ρ + 2 ) £ c v c m s ( α ρ + 1 , α ρ + 2 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 2 , α θ φ ( α ρ + 2 , α ρ + 3 ) ) £ c v c m s ( α ρ + 2 , α θ ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 2 , α θ φ ( α ρ + 3 , α θ ) ) £ c v c m s ( α ρ + 2 , α θ ) i 3 . . . i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + l = ρ + 1 θ 2 j = ρ + 1 l φ ( α j , α θ ) φ ( α l , α l + 1 ) £ c v c m s ( α l , α l + 1 ) + k = ρ + 1 θ 1 φ ( α k , α θ ) £ c v c m s ( α θ 1 , α θ ) i 3 φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 2 j = ρ + 1 l φ ( α j , α θ ) φ ( α l , α l + 1 ) l £ c v c m s ( α 0 , α 1 ) + k = ρ + 1 θ 1 φ ( α k , α θ ) θ 1 £ c v c m s ( α 0 , α 1 ) i 3 φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 2 j = ρ + 1 l φ ( α j , α θ ) φ ( α l , α l + 1 ) l £ c v c m s ( α 0 , α 1 ) + k = ρ + 1 θ 1 φ ( α k , α θ ) θ 1 φ ( α θ 1 , α θ ) £ c v c m s ( α 0 , α 1 ) = φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = n + 1 θ 1 j = ρ + 1 l φ ( α j , α θ ) φ ( α l , α l + 1 ) l £ c v c m s ( α 0 , α 1 ) i 3 φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 1 j = ρ + 1 l φ ( α j , α θ ) φ ( α l , α l + 1 ) l £ c v c m s ( α 0 , α 1 ) .
Also, φ ( α , g ) 1 . Let
S τ = l = 0 τ j = 0 l φ ( α j , α θ ) φ ( α l , α l + 1 ) l .
Hence we have,
£ c v c m s ( α ρ , α θ ) i 3 £ c v c m s ( α 0 , α 1 ) [ ρ φ ( α ρ , α ρ + 1 ) + ( S θ 1 , S ρ ) ] .
By (2) and applying the ratio test, we obtain that lim θ , ρ S ρ exists, so the real sequence { S ρ } is Cauchy.
As θ , ρ , we obtain
lim θ , ρ £ c v c m s ( α ρ , α θ ) = 0 .
Hence, { α ρ } is a Cauchy sequence in the complete Tcvcms ( , £ c v c m s ) ; and { α ρ } converges to a α * . By Lemma 1, { α ρ } has a unique limit. Moreover, from Lemma 2, we obtain
α * = lim ρ α ρ + 1 = lim ρ Γ α ρ = Γ ( lim ρ α ρ ) = Γ α * .
Let α * , g * Fix Γ . Then,
£ c v c m s ( α * , g * ) = £ c v c m s ( Γ α * , Γ g * ) i 3 φ £ c v c m s ( α * , g * ) .
Therefore, £ c v c m s ( α * , g * ) = 0 , then α * = g * . Hence, Γ has a fixed point. □
Theorem 2.
Let ( , £ c v c m s ) be a complete Tcvcms and Γ : be a mapping such that
£ c v c m s ( Γ α , Γ g ) i 3 φ £ c v c m s ( α , g ) ,
for all α , g , where 0 < < 1 . For α 0 , we denote α ρ = Γ ρ α 0 . Suppose that
max θ 1 lim l φ ( α l + 1 , α l + 2 ) φ ( α l , α l + 1 ) φ ( α l + 1 , α θ ) < 1 .
In addition, for every α , the limits
lim ρ φ ( α ρ , α ) and lim ρ φ ( α , α ρ ) exist and are finite .
Then, Γ has a UFP.
Proof. 
From Theorem 1 and using Lemma 2, we can find a Cauchy sequence { α ρ } in the complete Tcvcms ( , £ c v c m s ) . Then, the sequence { α ρ } converges to a α * . Therefore,
£ c v c m s ( α * , α ρ + 1 ) i 3 φ ( α * , α ρ ) £ c v c m s ( α * , α ρ ) + φ ( α ρ , α ρ + 1 ) £ c v c m s ( α * , α ρ + 1 ) .
Using (7), (8) and (18), we obtain
lim ρ £ c v c m s ( α * , α ρ + 1 ) = 0 .
Using the triangular inequality and (6), we have
£ c v c m s ( α * , Γ α * ) i 3 φ ( α * , α ρ + 1 ) £ c v c m s ( α * , α ρ + 1 ) + φ ( α ρ + 1 , Γ α * ) £ c v c m s ( α ρ + 1 , Γ α * ) i 3 φ ( α * , α ρ + 1 ) £ c v c m s ( α * , α ρ + 1 ) + φ ( α ρ + 1 , Γ α * ) £ c v c m s ( α ρ + 1 , Γ α * ) .
Taking the limit ρ and by (8) and (19), we obtain £ c v c m s ( α * , Γ α * ) = 0 . By Lemma 1, the sequence { α ρ } converges uniquely to α * . □
Example 1.
Consider = { 0 , 1 , 2 } and let £ c v c m s : × C 3 be a symmetric metric given by
£ c v c m s ( α , α ) = 0 , for each α
and
£ c v c m s ( 0 , 1 ) = 1 + i 3 , £ c v c m s ( 0 , 2 ) = 4 + 4 i 3 , £ c v c m s ( 1 , 2 ) = 1 + i 3 .
Define φ : × [ 1 , ) by
φ ( 0 , 0 ) = 2 , φ ( 0 , 1 ) = 3 2 , φ ( 0 , 2 ) = 4 3 , φ ( 1 , 1 ) = 4 3 , φ ( 1 , 2 ) = 5 4 , φ ( 2 , 2 ) = 6 5 .
Clearly, it is a Tcvcms.
Define the self map Γ on by Γ ( 0 ) = 0 , Γ ( 1 ) = 0 , Γ ( 2 ) = 0 .
Choosing = 2 5 . Then,
Case 1.
If α = g = 0 , α = g = 1 , α = g = 2 , then the result is obvious.
Case 2.
If α = 0 , g = 1 , we have
£ c v c m s ( Γ α , Γ g ) = £ c v c m s ( Γ 0 , Γ 1 ) = £ c v c m s ( 2 , 2 ) = 0 i 3 2 5 ( 1 + i 3 ) = ( £ c v c m s ( 0 , 1 ) ) = ( £ c v c m s ( α , g ) ) .
Case 3.
If α = 0 , g = 2 , we have
£ c v c m s ( Γ α , Γ g ) = £ c v c m s ( Γ 0 , Γ 1 ) = £ c v c m s ( 2 , 2 ) = 0 i 3 2 5 ( 4 + 4 i 3 ) = ( £ c v c m s ( 0 , 2 ) ) = ( £ c v c m s ( α , g ) ) .
Case 4.
If α = 1 , g = 2 , we have
£ c v c m s ( Γ α , Γ g ) = £ c v c m s ( Γ 1 , Γ 2 ) = £ c v c m s ( 2 , 2 ) = 0 i 3 2 5 ( 1 + i 3 ) = ( £ c v c m s ( 1 , 2 ) ) = ( £ c v c m s ( α , g ) ) .
Therefore, all hypotheses of Theorem 2 are satisfied. Hence, Γ has a UFP, which is α * = 0 .
Next, we establish the fixed point result using Kannan-type contraction mapping.
Theorem 3.
Let ( , £ c v c m s ) be a complete Tcvcms and Γ : be a continuous mapping such that
£ c v c m s ( Γ α , Γ g ) i 3 γ ( £ c v c m s ( α , Γ α ) + ( £ c v c m s ( g , Γ g ) ) ,
for all α , g , where 0 γ < 1 2 . For α 0 , we denote α ρ = Γ ρ α 0 . Suppose that
max θ 1 lim l φ ( α l + 1 , α l + 2 ) φ ( α l , α l + 1 ) φ ( α l + 1 , α θ ) < 1 , where = γ 1 γ .
Furthermore, for every α , if the limits
lim ρ φ ( α ρ , α ) and lim ρ φ ( α , α ρ ) exist and are finite ,
then Γ has a UFP.
Proof. 
For α 0 , consider a sequence { α ρ = Γ ρ α 0 } . If there exists α 0 N for which α ρ 0 + 1 = α ρ 0 , then Γ α ρ 0 = α ρ 0 , and we are done.
Let us assume that α ρ + 1 α ρ for all ρ N . By using (1), we obtain
£ c v c m s ( α ρ , α ρ + 1 ) = £ c v c m s ( Γ α ρ 1 , Γ α ρ ) i 3 γ ( £ κ ( α ρ 1 , Γ α ρ 1 ) + £ κ ( α ρ , Γ α ρ ) ) = γ ( £ c v c m s ( α ρ 1 , α ρ ) + £ c v c m s ( α ρ , α ρ + 1 ) ) ,
which implies £ c v c m s ( α ρ , α ρ + 1 ) i 3 γ 1 γ £ c v c m s ( α ρ 1 , α ρ ) = £ c v c m s ( α ρ 1 , α ρ ) . Similarly,
£ c v c m s ( α ρ 1 , α ρ ) = £ c v c m s ( Γ α ρ 2 , Γ α ρ 1 ) i 3 γ ( £ c v c m s ( α ρ 2 , Γ α ρ 2 ) + £ c v c m s ( α ρ 1 , Γ α ρ 1 ) ) = γ ( £ c v c m s ( α ρ 2 , α ρ 1 ) + £ c v c m s ( α ρ 1 , α ρ ) ) ,
which implies £ c v c m s ( α ρ 1 , α ρ ) i 3 γ 1 γ £ c v c m s ( α ρ 2 , α ρ 1 ) = £ c v c m s ( α ρ 2 , α ρ 1 ) .
Continuing in the same way, we have
£ c v c m s ( α ρ , α ρ + 1 ) i 3 £ c v c m s ( α ρ 1 , α ρ ) i 3 2 £ c v c m s ( α ρ 2 , α ρ 1 ) i 3 . . . . i 3 ρ £ c v c m s ( α 0 , α 1 ) .
Thus, £ c v c m s ( α ρ , α ρ + 1 ) i 3 ρ £ c v c m s ( α 0 , α 1 ) for all ρ 0 . For all ρ < θ , where ρ and θ are natural numbers, we have
£ c v c m s ( α ρ , α θ ) i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + φ ( α ρ + 1 , α θ ) £ c v c m s ( α ρ + 1 , α θ ) i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 1 , α ρ + 2 ) £ c v c m s ( α ρ + 1 , α ρ + 2 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 2 , α θ ) £ c v c m s ( α ρ + 2 , α θ ) i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 1 , α ρ + 2 ) £ c v c m s ( α ρ + 1 , α ρ + 2 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 2 , α θ ) φ ( α ρ + 2 , α ρ + 3 ) £ c v c m s ( α ρ + 2 , α ρ + 3 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 2 , α θ ) φ ( α ρ + 3 , α θ ) £ c v c m s ( α ρ + 3 , α θ ) i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) l = ρ + 1 θ 2 j = ρ + 1 l φ ( α j , α θ ) φ ( α l , α l + 1 ) £ c v c m s ( α l , α l + 1 ) + k = ρ + 1 θ 1 φ ( α k , α θ ) £ c v c m s ( α θ 1 , α θ ) i 3 φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 2 j = ρ + 1 l φ ( α j , α θ ) φ ( α l , α l + 1 ) l £ c v c m s ( α l , α l + 1 ) + k = ρ + 1 θ 1 φ ( α k , α θ ) θ 1 £ c v c m s ( α 0 , α 1 ) i 3 φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 2 j = ρ + 1 l φ ( α j , α θ ) φ ( α l , α l + 1 ) l £ c v c m s ( α l , α l + 1 ) + k = ρ + 1 θ 1 φ ( α k , α θ ) φ ( α θ 1 , α θ ) £ c v c m s ( α 0 , α 1 ) = φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 1 j = ρ + 1 l φ ( α j , α θ ) φ ( α l , α l + 1 ) l £ c v c m s ( α l , α l + 1 ) i 3 φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 1 j = ρ + 1 l α ( α j , α θ ) φ ( α l , α l + 1 ) l £ c v c m s ( α l , α l + 1 ) .
Moreover, φ ( α , g ) 1 . Let
S τ = l = 0 τ j = 0 l φ ( α j , α θ ) φ ( α l , α l + 1 ) l .
Hence, we have
£ c v c m s ( α ρ , α θ ) i 3 £ c v c m s ( α 0 , α 1 ) [ ρ φ ( α ρ , α ρ + 1 ) + ( S θ 1 , S ρ ) ] .
Applying the ratio test, we obtain that lim θ , ρ S ρ exists, so the sequence { S ρ } is a Cauchy sequence. As θ , ρ , we obtain
lim θ , ρ £ c v c m s ( α ρ , α θ ) = 0 .
Thus, { α ρ } is a Cauchy sequence in the complete Tcvcms ( , £ c v c m s ) . Accordingly, { α ρ } converges to some α * . By Lemma 1, { α ρ } has a unique limit. By Lemma 2, we obtain
α * = lim ρ α ρ + 1 = lim ρ Γ α ρ = Γ ( lim ρ α ρ ) = Γ α * .
Let α * , g * Fix Γ. Then,
£ c v c m s ( α * , g * ) = £ c v c m s ( Γ α * , Γ g * ) i 3 γ [ £ c v c m s ( α * , Γ α * ) + £ c v c m s ( g * , Γ g * ) ] i 3 γ [ £ c v c m s ( α * , α * ) + £ c v c m s ( g * , g * ) ] = 0 .
Therefore, £ c v c m s ( α * , g * ) = 0 , then α * = g * . Hence, Γ has a UFP. □
Theorem 4.
Let ( , £ c v c m s ) be a complete Tcvcms and Γ : be a mapping such that
£ c v c m s ( Γ α , Γ g ) i 3 γ ( £ c v c m s ( α , Γ α ) + £ c v c m s ( g , Γ g ) )
for all α , g where 0 γ < 1 2 . For α 0 , we denote α ρ = Γ ρ α 0 . Suppose that
max θ 1 lim l φ ( α l + 1 , α l + 2 ) φ ( α l , α l + 1 ) φ ( α l + 1 , α θ ) < 1 , where = γ 1 γ .
In addition, assume for every α that the limits
lim ρ φ ( α ρ , α ) and lim ρ φ ( α , α ρ ) exist and are finite .
Then, Γ has a UFP.
Proof. 
From Theorem 3 and using Lemma 2, we can find a Cauchy sequence { α ρ } in the complete Tcvcms ( , £ c v c m s ) . The sequence { α ρ } converges to a α * . Then
£ c v c m s ( α * , α ρ + 1 ) i 3 φ ( α * , α ρ ) £ c v c m s ( α * , α ρ ) + φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 )
Using (2), (3) and (18), we deduce
lim ρ α c v c m s ( α * , α ρ + 1 ) = 0 .
Using the triangular inequality and (1), we obtain
£ c v c m s ( α * , Γ α * ) i 3 φ ( α * , α ρ + 1 ) £ c v c m s ( α * , α ρ + 1 ) + φ ( α ρ + 1 , Γ α * ) £ c v c m s ( α ρ + 1 , Γ α * ) i 3 φ ( α * , α ρ + 1 ) £ c v c m s ( α * , α ρ + 1 ) + φ ( α ρ + 1 , Γ α * ) [ γ ( £ c v c m s ( α ρ , α ρ + 1 ) + £ c v c m s ( α * , Γ α * ) ) ] .
As ρ and by (3) and (19), we deduce that £ c v c m s ( α * , Γ α * ) = 0 . By Lemma 1, the sequence { α ρ } converges uniquely at the point α * . □
Example 2.
Let us consider = { 0 , 1 , 2 } , and let £ c v c m s : × C 3 be a symmetric metric given as follows
£ c v c m s ( α , α ) = 0 for each α
and
£ c v c m s ( 0 , 1 ) = 1 + i 3 , £ c v c m s ( 0 , 2 ) = 4 + 4 i 3 , £ c v c m s ( 1 , 2 ) = 1 + i 3 .
Define φ : × [ 1 , ) by
φ ( 0 , 0 ) = 5 , φ ( 0 , 1 ) = 3 , φ ( 0 , 2 ) = 7 3 , φ ( 1 , 1 ) = 7 3 , φ ( 1 , 2 ) = 2 , φ ( 2 , 2 ) = 9 5 .
Define the self-map Γ on by Γ ( 0 ) = Γ ( 1 ) = Γ ( 2 ) = 2 .
Choosing γ = 2 5 . Then,
Case 1.
If α = g = 0 , α = g = 1 , α = g = 2 , then the result is obvious.
Case 2.
If α = 0 , g = 1 , we have
£ c v c m s ( Γ α , Γ g ) = £ c v c m s ( Γ 0 , Γ 1 ) = £ c v c m s ( 2 , 2 ) = 0 i 3 10 5 ( 1 + i 3 ) = 2 5 ( 4 + 4 i 3 + ( 1 + i 3 ) ) = γ ( £ c v c m s ( 0 , 2 ) + £ c v c m s ( 1 , 2 ) ) = γ ( £ c v c m s ( α , Γ α ) + £ c v c m s ( g , Γ g ) ) .
Case 3.
If α = 0 , g = 2 , we have
£ c v c m s ( Γ α , Γ g ) = £ c v c m s ( Γ 0 , Γ 2 ) = £ c v c m s ( 2 , 2 ) = 0 i 3 8 5 ( 1 + i 3 ) = 2 5 ( 4 + 4 i 3 + 0 ) = γ ( £ c v c m s ( 0 , 2 ) + £ c v c m s ( 2 , 2 ) ) = γ ( £ c v c m s ( α , Γ α ) + £ c v c m s ( g , Γ g ) ) .
Case 4.
If α = 1 , g = 2 , we have
£ c v c m s ( Γ α , Γ g ) = £ c v c m s ( Γ 1 , Γ 2 ) = £ c v c m s ( 2 , 2 ) = 0 i 3 2 5 ( 1 + i 3 ) = 2 5 ( ( 1 + i 3 ) + 0 ) = γ ( £ c v c m s ( 1 , 2 ) + £ c v c m s ( 2 , 2 ) ) = γ ( £ c v c m s ( α , Γ α ) + £ c v c m s ( g , Γ g ) ) .
Then, all hypotheses of Theorem 4 are satisfied. Hence, T has a UFP, which is α * = 2 .
Finally, we prove the fixed point result using Fisher-type contraction mapping.
Theorem 5.
Let ( , £ c v c m s ) be a complete Tcvcms and Γ : be continuous mapping such that
£ c v c m s ( Γ α , Γ g ) i 3 ϖ £ c v c m s ( α , g ) + ς £ c v c m s ( α , Γ α ) £ c v c m s ( g , Γ g ) 1 + £ c v c m s ( α , g ) ,
for all α , g , where ϖ , ς [ 0 , 1 ) such that μ = ϖ 1 ς < 1 . For α 0 we denote α ρ = Γ ρ α 0 . Suppose that
max θ 1 lim i 3 φ ( α i 3 + 1 , α i 3 + 2 ) φ ( α i 3 , α i 3 + 1 ) φ ( α i 3 + 1 , α θ ) < 1 μ .
In addition, suppose that for every x the limits
lim ρ φ ( α ρ , α ) and lim ρ φ ( α , α ρ ) exist and are finite ,
then Γ has a UFP.
Proof. 
For α 0 , consider the sequence { α ρ = Γ ρ α 0 } . If there exists α 0 N for which α ρ 0 + 1 = α ρ 0 , then Γ α ρ 0 = α ρ 0 , and the result is trivial.
Suppose not. Let us assume that α ρ + 1 α ρ for all ρ N . By using (1), we obtain
£ c v c m s ( α ρ , α ρ + 1 ) = £ c v c m s ( Γ α ρ 1 , Γ α ρ ) i 3 ϖ £ c v c m s ( α ρ 1 , α ρ ) + ς £ c v c m s ( α ρ 1 , Γ α ρ ) £ c v c m s ( α ρ , Γ α ρ ) 1 + £ c v c m s ( α ρ 1 , α ρ ) = ϖ £ c v c m s ( α ρ 1 , α ρ ) + ς £ c v c m s ( α ρ 1 , α ρ ) £ c v c m s ( x ρ , α ρ ) 1 + £ c v c m s ( α ρ 1 , α ρ ) i 3 £ c v c m s ( x ρ 1 , α ρ ) + ς £ c v c m s ( α ρ , α ρ + 1 ) ,
which implies
£ c v c m s ( α ρ , α ρ + 1 ) i 3 ϖ 1 ς £ c v c m s ( α ρ 1 , α ρ ) = μ £ c v c m s ( α ρ 1 , α ρ ) .
Similarly,
£ c v c m s ( α ρ 1 , α ρ ) = £ c v c m s ( Γ α ρ 2 , Γ α ρ 1 ) i 3 ϖ £ c v c m s ( α ρ 2 , α ρ 1 ) + ς £ c v c m s ( α ρ 2 , Γ α ρ 2 ) £ c v c m s ( α ρ 1 , Γ α ρ 1 ) 1 + £ c v c m s ( α ρ 2 , α ρ 1 ) = ϖ £ c v c m s ( α ρ 2 , α ρ 1 ) + ς £ c v c m s ( α ρ 2 , α ρ 1 ) £ c v c m s ( α ρ 1 , α ρ ) 1 + £ c v c m s ( α ρ 2 , α ρ 1 ) i 3 ϖ £ c v c m s ( α ρ 2 , α ρ 1 ) + ς £ c v c m s ( α ρ 1 , α ρ ) ,
which implies
£ c v c m s ( α ρ 1 , α ρ ) i 3 ϖ 1 ς £ c v c m s ( α ρ 2 , α ρ 1 ) = μ ( α ρ 2 , α ρ 1 ) .
Continuing the same way, we have
£ c v c m s ( α ρ , α ρ + 1 ) i 3 μ £ c v c m s ( α ρ 1 , α ρ ) i 3 μ 2 £ c v c m s ( α ρ 2 , α ρ 1 ) i 3 . . . . i 3 μ ρ £ c v c m s ( α 0 , α 1 ) .
Thus, £ c v c m s ( α ρ , α ρ + 1 ) i 3 μ ρ £ c v c m s ( α 0 , α 1 ) for all ρ 0 . For all ρ < θ , where ρ and θ are natural numbers, we have
£ c v c m s ( α ρ , α θ ) i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + φ ( α ρ + 1 , α θ ) £ c v c m s ( α ρ + 1 , α θ ) i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 1 , α ρ + 2 ) £ c v c m s ( α ρ + 1 , α ρ + 2 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 2 , α θ ) £ c v c m s ( α ρ + 2 , α θ ) i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 1 , α ρ + 2 ) £ c v c m s ( α ρ + 1 , α ρ + 2 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 2 , α θ ) φ ( α ρ + 2 , α ρ + 3 ) £ c v c m s ( α ρ + 2 , α ρ + 3 ) + φ ( α ρ + 1 , α θ ) φ ( α ρ + 2 , α θ ) φ ( α ρ + 3 , α θ ) £ c v c m s ( α ρ + 3 , α θ ) i 3 . . . i 3 φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) + l = ρ + 1 θ 2 ( j = ρ + 1 l φ ( α j , α θ ) ) φ ( α l , α l + 1 ) £ c v c m s ( α l , α l + 1 ) + k = ρ + 1 θ 1 φ ( α k , α θ ) £ c v c m s ( α θ 1 , α θ ) i 3 φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 2 ( j = ρ + 1 l φ ( α j , α θ ) ) φ ( α l , α l + 1 ) l £ c v c m s ( α 0 , α 1 ) + k = ρ + 1 θ 1 φ ( α k , α θ ) θ 1 £ c v c m s ( α 0 , α 1 ) i 3 φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 2 ( j = ρ + 1 l φ ( α j , α θ ) ) φ ( α l , α l + 1 ) l £ c v c m s ( α 0 , α 1 ) + ( k = ρ + 1 θ 1 φ ( α k , α θ ) ) θ 1 φ ( α θ 1 , α θ ) £ c v c m s ( α 0 , α 1 ) = φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 1 ( j = ρ + 1 l φ ( α j , α θ ) ) φ ( α l , α l + 1 ) l £ c v c m s ( α 0 , α 1 ) i 3 φ ( α ρ , α ρ + 1 ) ρ £ c v c m s ( α 0 , α 1 ) + l = ρ + 1 θ 1 ( j = 0 l φ ( α j , α θ ) ) φ ( α l , α l + 1 ) l £ c v c m s ( α 0 , α 1 ) .
Moreover, φ ( α , g ) 1 . Let
S τ = l = 0 τ ( j = 0 l φ ( α j , α θ ) ) φ ( α l , α l + 1 ) l .
Hence, we have
£ c v c m s ( α ρ , α θ ) i 3 £ c v c m s ( α 0 , α 1 ) [ μ ρ φ ( α ρ , α ρ + 1 ) + ( S θ 1 , S ρ ) ] .
By using the ratio test, we obtain that lim θ , ρ S ρ exists and, hence, the real sequence { S ρ } is a Cauchy sequence. As θ , ρ , we deduce that
lim θ , ρ £ c v c m s ( α ρ , α θ ) = 0 .
Then, { α ρ } is a Cauchy sequence in the complete Tcvcms   ( , £ c v c m s ) . Therefore, the sequence { α ρ } converges to an α * . By Lemma 1, { α ρ } has a unique limit. By Lemma 2, we obtain
α * = lim ρ α ρ + 1 = lim ρ Γ α ρ = Γ ( lim ρ α ρ ) = Γ α * .
Let α * , g * Fix Γ be two fixed points of Γ. Then,
£ c v c m s ( Γ α * , Γ g * ) i 3 ϖ £ c v c m s ( α * , g * ) + ς £ c v c m s ( α * , Γ g * ) £ c v c m s ( α * , Γ g * ) 1 + £ c v c m s ( α * , g * ) i 3 ϖ £ c v c m s ( α * , g * ) + ς £ c v c m s ( α * , α * ) £ c v c m s ( g * , g * ) 1 + £ c v c m s ( α * , g * ) i 3 ϖ £ c v c m s ( α * , g * ) .
Therefore, £ c v c m s ( α * , g * ) = 0 ; then α * = g * . Hence, Γ has a UFP. □
If we omit the continuous condition in Theorem 5, we obtain the following results.
Theorem 6.
Let ( , £ c v c m s ) be a complete Tcvcms and Γ : be a mapping such that
£ c v c m s ( Γ α , Γ g ) i 3 ϖ £ c v c m s ( α , g ) + ς £ c v c m s ( α , Γ α ) £ c v c m s ( g , Γ g ) 1 + £ c v c m s ( α , g ) ,
for all α , g , where ϖ , ς [ 0 , 1 ) such that μ = ϖ 1 ς < 1 . For α 0 , we denote α ρ = Γ ρ α 0 . Suppose that
max θ 1 lim i 3 φ ( α i 3 + 1 , α i 3 + 2 ) φ ( α i 3 3 , α i 3 + 1 ) φ ( α i 3 + 1 , α θ ) < 1 μ .
In addition, for every α , we have that
lim ρ φ ( α ρ , α ) and lim ρ ( α , α ρ ) exist and are finite .
Then, Γ has a UFP.
Proof. 
Using Theorem 5 and by Lemma 2, we obtain a Cauchy sequence { α ρ } which converges to an α * . Then,
£ c v c m s ( α * , α ρ + 1 ) i 3 φ ( α * , α ρ ) £ c v c m s ( α * , α ρ ) + φ ( α ρ , α ρ + 1 ) £ c v c m s ( α ρ , α ρ + 1 ) .
Using (2), (3) and (23), we deduce that
lim ρ £ c v c m s ( α * , α ρ + 1 ) = 0 .
Using the triangular inequality and (1),
£ c v c m s ( α * , Γ α * ) i 3 φ ( α * , α ρ + 1 ) £ c v c m s ( α * , α ρ + 1 ) + φ ( α ρ + 1 , Γ * ) £ c v c m s ( α ρ + 1 , Γ α * ) i 3 φ ( α * , α ρ + 1 ) £ c v c m s ( α * , α ρ + 1 ) + φ ( α ρ + 1 , Γ * ) [ ϖ £ c v c m s ( α ρ , α * ) + ς £ c v c m s ( α ρ , Γ α ρ ) £ c v c m s ( α * , Γ α * ) 1 + £ c v c m s ( α ρ , α * ) ] = φ ( α * , α ρ + 1 ) £ c v c m s ( α * , α ρ + 1 ) + φ ( α ρ + 1 , Γ * ) [ ϖ £ c v c m s ( α ρ , α * ) + ς £ c v c m s ( α ρ , Γ α ρ ) £ c v c m s ( α * , Γ α * ) 1 + £ c v c m s ( α ρ , α * ) ] .
As ρ , and using (3) and (24), we deduce that £ c v c m s ( α * , Γ α * ) = 0 . By Lemma 1, the sequence { α ρ } converges uniquely at the point α * . □
Example 3.
Let us consider = { 0 , 1 , 2 } and let £ c v c m s : × C be a symmetric metric given as
£ c v c m s ( α , α ) = 0 for each α
and
£ c v c m s ( 0 , 1 ) = 1 + i 3 , £ c v c m s ( 0 , 2 ) = 4 + 4 i 3 , £ c v c m s ( 1 , 2 ) = 1 + i 3 .
Define φ : × [ 1 , ) by
φ ( 0 , 0 ) = 5 , φ ( 0 , 1 ) = 3 , φ ( 0 , 2 ) = 7 3 , φ ( 1 , 1 ) = 7 3 , φ ( 1 , 2 ) = 2 , φ ( 2 , 2 ) = 9 5 .
Clearly, ( , £ c v c m s ) is a Tcvcms. Define the self-map Γ on by Γ ( 0 ) = Γ ( 1 ) = Γ ( 2 ) = 1 .
If we choose ϖ = ς = 1 5 , we have
Case 1.
If α = g = 0 , α = g = 1 , α = g = 2 , we have £ c v c m s ( Γ α , Γ g ) = 0 .
Case 2.
If α = 0 , g = 1 , we have £ c v c m s ( Γ α , Γ g ) = 0 i 3 ϖ £ c v c m s ( α , g ) + ς £ c v c m s ( α , Γ α ) £ c v c m s ( g , Γ g ) 1 + £ c v c m s ( α , g ) .
Case 3.
If α = 0 , g = 2 , we have £ c v c m s ( Γ α , Γ g ) = 0 i 3 ϖ £ c v c m s ( α , g ) + ς £ c v c m s ( α , Γ α ) £ c v c m s ( g , Γ g ) 1 + £ c v c m s ( α , g ) .
Case 4.
If α = 1 , g = 2 , we have £ c v c m s ( Γ α , Γ g ) = 0 i 3 ϖ £ c v c m s ( α , g ) + ς £ c v c m s ( α , Γ α ) £ c v c m s ( g , Γ g ) 1 + £ c v c m s ( α , g ) .
Thus, all hypotheses of Theorem 6 are satisfied. Hence, Γ has a UFP, which is α * = 1 .

4. Application

Consider the integral equation
x ( τ ) = r ( τ ) + 0 τ ( τ , κ ) g ( κ , x ( κ ) ) α κ , τ [ 0 , 1 ] , x ( τ ) ,
where g ( τ , x ( τ ) ) : [ 0 , 1 ] × R R , r ( τ ) : [ 0 , 1 ] R are two bounded and continuous functions and : [ 0 , 1 ] × [ 0 , 1 ] [ 0 , ) is a function satisfying ( τ , · ) L 1 ( [ 0 , 1 ] ) for all τ [ 0 , 1 ] .
Theorem 7.
Let = C ( [ 0 , 1 ] , R ) = { F | F : [ 0 , 1 ] R is a continuous function } . Moreover, let Γ : be an operator of the form:
x ( τ ) = r ( τ ) + 0 τ ( τ , κ ) g ( κ , x ( κ ) ) α κ , τ [ 0 , 1 ] , x ( τ ) .
Suppose the following conditions hold:
(i)
The functions g ( τ , x ( τ ) ) : [ 0 , 1 ] × R R and r ( τ ) : [ 0 , 1 ] R are continuous.
(ii)
: [ 0 , 1 ] × [ 0 , 1 ] [ 0 , ) be a function with ( τ , · ) L 1 ( [ 0 , 1 ] ) τ [ 0 , 1 ] we have:
0 τ ( τ , κ ) α κ < 1 .
(iii)
| g ( τ , x ( τ ) ) g ( τ , y ( τ ) ) | 1 ω · g i 3 ω τ | x ( τ ) y ( τ ) | , for all x , y and ω ( 1 , 1 ] with ( 0 , 1 ) .
Then, the Equation (25) has a unique solution.
Proof. 
Let = C ( [ 0 , 1 ] , R ) and £ c v c m s : × C 3 be a tricomplex-valued controlled metric such that for every ω ( 1 , 1 ] , ( 0 , 1 )
£ c v c m s ( x , y ) = x = sup τ [ 0 , 1 ] | x ( τ ) | g i 3 ω τ
and define φ τ : × [ 1 , ) by
φ τ ( x , y ) = 1 , if x , y [ 0 , 1 ] , max { x ( τ ) , y ( τ ) } + ω , otherwise .
Clearly, ( , £ c v c m s ) is a complete Tcvcms. Now,
| Γ x ( τ ) Γ y ( τ ) | | 0 τ [ ( τ , κ ) g ( κ , x ( κ ) ) ( τ , κ ) g ( κ , y ( κ ) ) ] α κ | 0 τ | ( τ , κ ) [ g ( κ , x ( κ ) ) g ( κ , y ( κ ) ) ] | α κ 0 τ ( τ , κ ) α κ 0 τ | [ g ( κ , x ( κ ) ) g ( κ , y ( κ ) ) ] | α κ 1 τ · g i 3 ω τ 0 τ | x ( κ ) y ( κ ) | α κ 0 τ ( τ , κ ) α κ = g i 3 ω τ τ · g i 3 ω τ g i 3 ω τ 0 τ | x ( κ ) y ( τ ) | g i 3 ω t α κ 0 τ ( τ , κ ) g i 3 ω τ α κ .
Taking the supremum, we obtain
[ sup τ [ 0 , 1 ] | Γ x ( τ ) Γ y ( τ ) | g i 3 ω τ ] 1 ω [ sup τ [ 0 , 1 ] | Γ x ( τ ) Γ y ( τ ) | g i 3 ω τ ] · 0 τ sup τ [ 0 , 1 ] ( τ , κ ) g i 3 ω τ α κ .
Using hypothesis (ii), we have
£ c v c m s ( Γ x τ ) = Γ x Γ y 1 ω x y = 1 ω £ c v c m s ( x , y ) .
Then, for 0 < δ = 1 ω < 1 , all the hypotheses of Theorem 1 are satisfied. Hence, the Equation (25) has a unique solution. □

5. Conclusions

In this paper, the concept of Tcvcms was introduced, and we proved some fixed point results for Banach-, Kannan- and Fisher-type contractions. The established results are supported with nontrivial examples. We also provided an application to find the analytical solution of an integral equation. It will be an open problem to establish fixed point results using Meir–Keeler- and Presic-type contractions in the setting of Tcvcms. Recently, Khalehoghli et al. [20] introduced R-metric spaces and obtained a generalization of Banach fixed point theorem. It is an interesting open problem to study the R-tricomplex-valued controlled metric spaces and obtain fixed point results on complete R-tricomplex-valued controlled metric spaces.

Author Contributions

Investigation: G.M., R.R. and A.J.G.; methodology: R.R. and G.M.; project administration: R.R. and S.R.; software: A.J.G. and O.A.A.A.; supervision: R.R. and S.R.; writing—original draft: G.M. and R.R.; writing—review and editing: R.R., O.A.A.A. and S.R. All authors have read and agreed to the published version of the manuscript.

Funding

The authors extend their appreciation to the Deputyship for Research & Innovation, Ministry of Education in Saudi Arabia for funding this research work through the project number IF2/PSAU/2022/01/22334.

Data Availability Statement

Not applicable.

Acknowledgments

The authors convey their sincere thanks to the anonymous reviewers for their valuable comments which helped in bringing the manuscript to its present form.

Conflicts of Interest

The authors declare no conflict of interest.

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Ramaswamy, R.; Mani, G.; Gnanaprakasam, A.J.; Abdelnaby, O.A.A.; Radenović, S. Solving an Integral Equation via Tricomplex-Valued Controlled Metric Spaces. Axioms 2023, 12, 56. https://doi.org/10.3390/axioms12010056

AMA Style

Ramaswamy R, Mani G, Gnanaprakasam AJ, Abdelnaby OAA, Radenović S. Solving an Integral Equation via Tricomplex-Valued Controlled Metric Spaces. Axioms. 2023; 12(1):56. https://doi.org/10.3390/axioms12010056

Chicago/Turabian Style

Ramaswamy, Rajagopalan, Gunaseelan Mani, Arul Joseph Gnanaprakasam, Ola A. Ashour Abdelnaby, and Stojan Radenović. 2023. "Solving an Integral Equation via Tricomplex-Valued Controlled Metric Spaces" Axioms 12, no. 1: 56. https://doi.org/10.3390/axioms12010056

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